= Solution
{c}
Put $Q=Q_r(x)$ and $v_Q=|Q|^{-1}\int_Qv$. The <fundamental theorem of calculus along a line segment>, followed by averaging, gives
$$
|v(y)-v_Q|\leq\frac1{r^n}\int_Q\int_0^1|Dv(y+t(z-y))|\,|z-y|\,dt\,dz.
$$
For fixed $t$, use the <change of variables formula> $w=y+t(z-y)$. Its <Jacobian determinant> is $t^n$. The transformed domain $Q_t=y+t(Q-y)$ lies in $Q$, because a cube is a <convex set>. Moreover, $w\in Q_t$ implies $|w-y|\leq\sqrt n\,rt$. Thus <Tonelli theorem> gives
$$
\begin{aligned}
|v(y)-v_Q|
&\leq\frac1{r^n}\int_Q |Dv(w)|\,|w-y|
\int_{|w-y|/(\sqrt n r)}^1t^{-n-1}dt\,dw\\
&\leq C_n\int_Q\frac{|Dv(w)|}{|w-y|^{n-1}}\,dw.
\end{aligned}
$$
The diagonal $w=y$ is a null set. The <Holder inequality> with $p'=p/(p-1)$ now applies: $p>n$ implies $(n-1)p'<n$, and integration in <spherical coordinates> bounds the kernel by
$$
\left(\int_Q|w-y|^{-(n-1)p'}dw\right)^{1/p'}
\leq C_{n,p}r^{n/p'-(n-1)}=C_{n,p}r^{1-n/p}.
$$
This proves the <Morrey inequality on a cube>, uniformly even when $y$ approaches its boundary:
$$
\boxed{|v_Q-v(y)|\leq C_1r^{1-n/p}\|Dv\|_{L^p(Q)}.}
$$
Apply this at both $y$ and the center $x$ and use the triangle inequality to obtain
$$
\boxed{|v(y)-v(x)|\leq C_2r^{1-n/p}\|Dv\|_{L^p(Q)},\qquad C_2=2C_1.}
$$
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